Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let me first explain the background of my question.

As is well known, the group $SO(n+1)$ acts transitively on the sphere $S^n$, and the stabilizer is the group $SO(n)$, so that we get a fibration sequence $$ SO(n) \to SO(n+1) \to S^n.$$

Indeed this fibration is a principal-$SO(n)$-bundle and can actually be identified with the frame bundle of the tangent bundle of $S^n$, i.e., the tangent bundle $TS^n$ is the associated bundle to this principal-$SO(n)$-bundle along the tautological representation of $SO(n)$ on $\mathbb{R}^n$.

Now let us consider the special case $n = 4$. Here we get a principal-$SO(3)$-bundle $$ SO(3) \to SO(4) \to S^3.$$ But we can identify the $S^3$ with $Sp(1)$, the symplectic group acting on the quaternions. Via this identification $S^3$ sits canonically in $SO(4)$, i.e. we have an embedding of topological groups $S^3 \cong Sp(1) \to SO(4)$.

The standard theory of principal-bundles hence tells us that the above bundle splits (topologically) which means that there is a homeomorphism $$ SO(4) \cong SO(3)\times S^3 $$ in particular there is a continuous map $SO(4) \to SO(3)$ (which is not a morphism of lie groups).

It is well known that the homotopy type $BSO(n)$ represents the functor of $n$-dimensional oriented vector bundles, and via the clutching function construction the group $SO(n)$ itself represents the functor of $n$-dimensional vector bundles over suspensions of spaces. Hence the above map $SO(4) \to SO(3)$ tells us that there is a natural transformation between the functor of rank $4$ vector bundles over suspensions to the functor of rank $3$ vector bundles over suspensions.

Does anybody know a geometric construction of this transformation which I only understand homotopy theoretically?

Any ideas or references are greatly appreciated.

share|improve this question
    
Just coincidences of Dynkin diagrams gives you that $SO(4)$ and $SO(3) \times SL(2)$ have the same double cover (namely, $SL(2) \times SL(2) = Spin(4)$). Not quite sure where to go from there since they're quotients by different $Z/2$ actions, and it's not clear to me why you'd get the same manifold either way. –  Noah Snyder Dec 22 '12 at 16:06
3  
@NoahSnyder You mean SU(2), not SL(2), right? (Same thing up to homotopy, but we might as well get it right since the question asks for a geometric interpretation.) –  David Speyer Dec 22 '12 at 17:22
    
I don't quite see where you're getting the map $Sp(1)\rightarrow SO(4)$ from. In general, nontrivial principal bundles have no sections. On the other hand, this particular bundle is trivial, but there are two different choices of trivialization which are both distinct homomorphisms and distinct from the homotopy perspective (though, they're similar enough that whatever answer you get for one will work for the other). –  Jason DeVito Dec 22 '12 at 17:48
1  
I see what you're saying - but there are two such inclusions. Anyway, I think I've found a paper which answers your question - I'll post an answer in a second. –  Jason DeVito Dec 22 '12 at 18:05
1  
@DavidSpeyer: Yes, SU(2), sadly too late to edit. –  Noah Snyder Dec 22 '12 at 19:20
show 1 more comment

1 Answer

up vote 4 down vote accepted

The double cover $\pi:S^3\times S^3\rightarrow SO(4)$ actually provides two homotopoically distinct maps from $S^3 = Sp(1)$ to $SO(4)$, given by restricting $\pi$ to either factor. (One can easily see that these two maps induce different maps on $\pi_3$, so are not homotopic).

The following theorem is contained in

K.Grove-W.Ziller, Lifting group actions and nonnegative curvature, Trans. Amer. Math. Soc. 363 (2011) 2865-2890.

Ziller has a freely accessible version here - see the middle of page $8$.

Theorem If $E\rightarrow M$ is a rank $4$ vector bundle over $M$ (where $M$ is a compact simply connected manifold.), then $\Lambda^2(E) = \Lambda^2(E)_+\oplus\Lambda^2(E)_-$ decomposes into self-dual and antiself dual forms. Then $\Lambda^2(E)_{\pm}$ are the two rank $3$ vector bundles over $M$ corresponding to the two sections above.

share|improve this answer
    
This sounds very interesting, thank you! But can you remind me what the involution on $\Lambda^2(E)$ is that gives the decomposition? Moreover where can I find an explicit formula for the map $\pi: S^3\times S^3 \to SO(4)$? –  mland Dec 22 '12 at 18:24
6  
The involution is the Hodge star, which in general is a map $\ast: \Lambda^k(E)\rightarrow \Lambda^{n-k}(E)$ (where $E$ has rank $n$) depending on the choice of inner product on each fiber. An explicit formula for the double cover $S^3\times S^3\rightarrow SO(4)$ comes from viewing everything as unit quaternions. It takes $(p,q)\in S^3\times S^3$ to the linear map $f_{(p,q)}$ on $\mathbb{R}^4 = \mathbb{H}$ given by $f_{(p,q)}(v) = pv\overline{q}$. –  Jason DeVito Dec 22 '12 at 18:31
    
Many thanks again! What I still find kind of confusing, is that I do not really see, where we use that this transformation should be given only on suspensions of spaces.. What do I miss here? –  mland Dec 23 '12 at 15:30
    
Well, assuming the usual space is connected (and I'm not exactly sure where this enters), it's suspension is simply connected, which is one of the hypothesis of their theorem. Also, I've been thinking about it and I know longer thing the decomposition comes (canonically) from the Hodge star. The issue is that the Hodge star requires a metric to define. On the other hand, I think that given the standard rep $V$ of $SU(2)\times SU(2)$, then $\Lambda^2 V$ has a unique nontrivial $SU(2)\times \{e\}$ invt. subspace as does $\{e\}\times SU(2)$ and the sum of these is $\Lambda^2 V$. –  Jason DeVito Dec 23 '12 at 15:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.